Mixing
Rank($U$) = Rank($U^2$) for group of units $U$
$begingroup$
I am reading the paper "Algebraic Integers on the Unit Circle" by Ryan C. Daileda (http://www.sciencedirect.com/science/article/pii/S0022314X05002027). I am confused about how he concludes that the rank of $U$ is equal to the rank of $U^2$ in the proof of Theorem 1.
Can anyone please explain?
(Edit to add the theorem and proof):
Let $K$ be an algebraic number field, $U$ be its group of units, $V$ be the subgroup of units of modulus $1$, and $R$ be the group of real units in $K$ $(R = U cap mathbb{R})$.
Then the theorem states: "Let $K$ be a number field closed under complex conjugation, and let $U, V,$ and $R$ be as above. Then $operatorname{rank}(V)+operatorname{rank}(R) = operatorname{rank}(U)$." The proof is:
"By hypothesis, if $u in U$, then $bar{u} in U$ as well. Therefore, $ubar{u} = |u|^2 in R$, and $u/bar{u} in V$. This means that $u^2 = |u|^2 frac{u}{bar{u}} in RV$.
Hence $U^2 subset RV subset U$, so that $operatorname{rank}(U) = operatorname{rank}(U^2)= operatorname{rank}(RV)$. It is clear that $Rcap V = {pm 1}$, giving $operatorname{rank}(RV) = operatorname{rank}(R) + operatorname{rank}(V)$, so we're done."
algebraic-number-theory abelian-groups
asked Jul 16 '14 at 23:49
user164793user164793
83
$endgroup$
add a comment |
$begingroup$
I am reading the paper "Algebraic Integers on the Unit Circle" by Ryan C. Daileda (http://www.sciencedirect.com/science/article/pii/S0022314X05002027). I am confused about how he concludes that the rank of $U$ is equal to the rank of $U^2$ in the proof of Theorem 1.
Can anyone please explain?
(Edit to add the theorem and proof):
Let $K$ be an algebraic number field, $U$ be its group of units, $V$ be the subgroup of units of modulus $1$, and $R$ be the group of real units in $K$ $(R = U cap mathbb{R})$.
Then the theorem states: "Let $K$ be a number field closed under complex conjugation, and let $U, V,$ and $R$ be as above. Then $operatorname{rank}(V)+operatorname{rank}(R) = operatorname{rank}(U)$." The proof is:
"By hypothesis, if $u in U$, then $bar{u} in U$ as well. Therefore, $ubar{u} = |u|^2 in R$, and $u/bar{u} in V$. This means that $u^2 = |u|^2 frac{u}{bar{u}} in RV$.
Hence $U^2 subset RV subset U$, so that $operatorname{rank}(U) = operatorname{rank}(U^2)= operatorname{rank}(RV)$. It is clear that $Rcap V = {pm 1}$, giving $operatorname{rank}(RV) = operatorname{rank}(R) + operatorname{rank}(V)$, so we're done."
algebraic-number-theory abelian-groups
asked Jul 16 '14 at 23:49
user164793user164793
83
$endgroup$
$begingroup$
I cannot access the paper from home: can you type the theorem? I suspect it is a finite index thing, depending on how you're defining $U^2$ (some do products, others do squares)
$endgroup$
– Adam Hughes
Jul 16 '14 at 23:52
$begingroup$
I believe $U^2$ means squares but it is not actually stated. It just seems that way from the proof.
$endgroup$
– user164793
Jul 17 '14 at 0:03
$begingroup$
It does indeed, I can tell from the proof. I've posted an answer.
$endgroup$
– Adam Hughes
Jul 17 '14 at 0:05
add a comment |
$begingroup$
I am reading the paper "Algebraic Integers on the Unit Circle" by Ryan C. Daileda (http://www.sciencedirect.com/science/article/pii/S0022314X05002027). I am confused about how he concludes that the rank of $U$ is equal to the rank of $U^2$ in the proof of Theorem 1.
Can anyone please explain?
(Edit to add the theorem and proof):
Let $K$ be an algebraic number field, $U$ be its group of units, $V$ be the subgroup of units of modulus $1$, and $R$ be the group of real units in $K$ $(R = U cap mathbb{R})$.
Then the theorem states: "Let $K$ be a number field closed under complex conjugation, and let $U, V,$ and $R$ be as above. Then $operatorname{rank}(V)+operatorname{rank}(R) = operatorname{rank}(U)$." The proof is:
"By hypothesis, if $u in U$, then $bar{u} in U$ as well. Therefore, $ubar{u} = |u|^2 in R$, and $u/bar{u} in V$. This means that $u^2 = |u|^2 frac{u}{bar{u}} in RV$.
Hence $U^2 subset RV subset U$, so that $operatorname{rank}(U) = operatorname{rank}(U^2)= operatorname{rank}(RV)$. It is clear that $Rcap V = {pm 1}$, giving $operatorname{rank}(RV) = operatorname{rank}(R) + operatorname{rank}(V)$, so we're done."
algebraic-number-theory abelian-groups
asked Jul 16 '14 at 23:49
user164793user164793
83
$endgroup$
I am reading the paper "Algebraic Integers on the Unit Circle" by Ryan C. Daileda (http://www.sciencedirect.com/science/article/pii/S0022314X05002027). I am confused about how he concludes that the rank of $U$ is equal to the rank of $U^2$ in the proof of Theorem 1.
Can anyone please explain?
(Edit to add the theorem and proof):
Let $K$ be an algebraic number field, $U$ be its group of units, $V$ be the subgroup of units of modulus $1$, and $R$ be the group of real units in $K$ $(R = U cap mathbb{R})$.
Then the theorem states: "Let $K$ be a number field closed under complex conjugation, and let $U, V,$ and $R$ be as above. Then $operatorname{rank}(V)+operatorname{rank}(R) = operatorname{rank}(U)$." The proof is:
"By hypothesis, if $u in U$, then $bar{u} in U$ as well. Therefore, $ubar{u} = |u|^2 in R$, and $u/bar{u} in V$. This means that $u^2 = |u|^2 frac{u}{bar{u}} in RV$.
Hence $U^2 subset RV subset U$, so that $operatorname{rank}(U) = operatorname{rank}(U^2)= operatorname{rank}(RV)$. It is clear that $Rcap V = {pm 1}$, giving $operatorname{rank}(RV) = operatorname{rank}(R) + operatorname{rank}(V)$, so we're done."
algebraic-number-theory abelian-groups
algebraic-number-theory abelian-groups
asked Jul 16 '14 at 23:49
user164793user164793
83
asked Jul 16 '14 at 23:49
user164793user164793
83
edited Jul 16 '14 at 23:57
user164793
asked Jul 16 '14 at 23:49
user164793user164793
83
asked Jul 16 '14 at 23:49
user164793user164793
83
asked Jul 16 '14 at 23:49
user164793user164793
83
83
$begingroup$
I cannot access the paper from home: can you type the theorem? I suspect it is a finite index thing, depending on how you're defining $U^2$ (some do products, others do squares)
$endgroup$
– Adam Hughes
Jul 16 '14 at 23:52
$begingroup$
I believe $U^2$ means squares but it is not actually stated. It just seems that way from the proof.
$endgroup$
– user164793
Jul 17 '14 at 0:03
$begingroup$
It does indeed, I can tell from the proof. I've posted an answer.
$endgroup$
– Adam Hughes
Jul 17 '14 at 0:05
add a comment |
$begingroup$
I cannot access the paper from home: can you type the theorem? I suspect it is a finite index thing, depending on how you're defining $U^2$ (some do products, others do squares)
$endgroup$
– Adam Hughes
Jul 16 '14 at 23:52
$begingroup$
I believe $U^2$ means squares but it is not actually stated. It just seems that way from the proof.
$endgroup$
– user164793
Jul 17 '14 at 0:03
$begingroup$
It does indeed, I can tell from the proof. I've posted an answer.
$endgroup$
– Adam Hughes
Jul 17 '14 at 0:05
$begingroup$
I cannot access the paper from home: can you type the theorem? I suspect it is a finite index thing, depending on how you're defining $U^2$ (some do products, others do squares)
$endgroup$
– Adam Hughes
Jul 16 '14 at 23:52
$begingroup$
I cannot access the paper from home: can you type the theorem? I suspect it is a finite index thing, depending on how you're defining $U^2$ (some do products, others do squares)
$endgroup$
– Adam Hughes
Jul 16 '14 at 23:52
$begingroup$
I believe $U^2$ means squares but it is not actually stated. It just seems that way from the proof.
$endgroup$
– user164793
Jul 17 '14 at 0:03
$begingroup$
I believe $U^2$ means squares but it is not actually stated. It just seems that way from the proof.
$endgroup$
– user164793
Jul 17 '14 at 0:03
$begingroup$
It does indeed, I can tell from the proof. I've posted an answer.
$endgroup$
– Adam Hughes
Jul 17 '14 at 0:05
$begingroup$
It does indeed, I can tell from the proof. I've posted an answer.
$endgroup$
– Adam Hughes
Jul 17 '14 at 0:05
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
All right, well first of all Dirichlet's unit theorem tells you that the unit group (modulo roots of unity, which are irrelevant for computing rank) is a finitely generated free group on $r+s-1$ generators with $r$ the number of real embeddings and $s$ the number of complex ones.
If we use additive notation
$$U/text{Tor}(U)cong bigoplus_{i=1}^{r+s-1}Bbb Z$$
it is easier to see that the squares here are a subgroup of index $2^{r+s-1}$ since $U^2/text{Tor}(U^2)$ is generated by squares of generators of the torsion-free part of $U$. But then if the subgroup is of finite index, they have the same rank.
Edit: In case you've not seen the proof of finite index explicitly, it's quite direct:
If $xin U^2$ and $x_1,ldots, x_{r+s-1}$ are generators for $U/text{Tor}(U)$ then $x=(x_1^{e_1}ldots x_{r+s-1}^{e_{r+s-1}})^2$ so that the powers are all even, hence $x_1^2,ldots, x_{r+s-1}^2$ generate $U^2/text{Tor}(U^2)$. Consider the projection map
$$U/text{Tor}(U)to U^2/text{Tor}(U^2)$$
which is given on the additive structure by the multiplication by $2$ map, you get an isomorphism
$$U^2/text{Tor}(U^2)cong (2Bbb Z)^{r+s-1}$$
for the image.
Edit 2: Let's just do this explicitly. In this particular case it's not even hard. You can see that $x_1^{2e_1}cdotldotscdot x_{r+s-1}^{2e_{r+s-1}}=1$ (i.e. the identity element of the group) implies that all the $e_i$ are zero because we know that this holds in $U$ which is a bigger group. Hence $x_1,ldots ,x_{r+s-1}$ are linearly independent elements of $U^2$ hence it has rank $ge r+s-1$, but it also has rank $le r+s-1$ because all elements of $U^2$ are also elements of $U$ and it has rank $n$.
I think what confused you is thinking that the things in the proof were intended to prove $text{rank}(U^2)=text{rank}(U)$, but in reality that's known from this argument via basic algebra. The real reason it's in there is because you want to trap the group $RV$ which is between them to compute its rank, which is part of what was being shown.
answered Jul 17 '14 at 0:04
Adam HughesAdam Hughes
32.1k83670
$endgroup$
$begingroup$
"But then if the subgroup is of finite index, they have the same rank." - is this a general fact about free abelian groups?
$endgroup$
– user164793
Jul 17 '14 at 0:20
$begingroup$
@user164793 Finitely generated ones? Yes. Easier way than my previous comment: since you're dealing with a PID, you can form the vector space spanned by the basis elements for $Bbb Z^n$, and then notice that modding out by a subspace of the same rank gives the trivial vector space by the definition of dimension.
$endgroup$
– Adam Hughes
Jul 17 '14 at 0:33
$begingroup$
I'm sorry, I still don't follow why the finite index of $U^2$ in $U$ allows us to conclude they are of equal rank. I'm fairly weak in algebra, if you haven't picked up on that :( I understand the rest.
$endgroup$
– user164793
Jul 17 '14 at 0:41
$begingroup$
@user164793 give me a moment to edit my post, I'll just do the proof, it's not hard.
$endgroup$
– Adam Hughes
Jul 17 '14 at 0:43
$begingroup$
Thank you very much, I understand now. I really appreciate you taking the time.
$endgroup$
– user164793
Jul 17 '14 at 0:49
|
show 2 more comments
$begingroup$
$U/U^2$ is a finitely generated (multiplicative) abelian group every element of which has order $2$. Thus it is finite.
answered Jul 17 '14 at 0:04
Rene SchipperusRene Schipperus
32.2k11960
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
All right, well first of all Dirichlet's unit theorem tells you that the unit group (modulo roots of unity, which are irrelevant for computing rank) is a finitely generated free group on $r+s-1$ generators with $r$ the number of real embeddings and $s$ the number of complex ones.
If we use additive notation
$$U/text{Tor}(U)cong bigoplus_{i=1}^{r+s-1}Bbb Z$$
it is easier to see that the squares here are a subgroup of index $2^{r+s-1}$ since $U^2/text{Tor}(U^2)$ is generated by squares of generators of the torsion-free part of $U$. But then if the subgroup is of finite index, they have the same rank.
Edit: In case you've not seen the proof of finite index explicitly, it's quite direct:
If $xin U^2$ and $x_1,ldots, x_{r+s-1}$ are generators for $U/text{Tor}(U)$ then $x=(x_1^{e_1}ldots x_{r+s-1}^{e_{r+s-1}})^2$ so that the powers are all even, hence $x_1^2,ldots, x_{r+s-1}^2$ generate $U^2/text{Tor}(U^2)$. Consider the projection map
$$U/text{Tor}(U)to U^2/text{Tor}(U^2)$$
which is given on the additive structure by the multiplication by $2$ map, you get an isomorphism
$$U^2/text{Tor}(U^2)cong (2Bbb Z)^{r+s-1}$$
for the image.
Edit 2: Let's just do this explicitly. In this particular case it's not even hard. You can see that $x_1^{2e_1}cdotldotscdot x_{r+s-1}^{2e_{r+s-1}}=1$ (i.e. the identity element of the group) implies that all the $e_i$ are zero because we know that this holds in $U$ which is a bigger group. Hence $x_1,ldots ,x_{r+s-1}$ are linearly independent elements of $U^2$ hence it has rank $ge r+s-1$, but it also has rank $le r+s-1$ because all elements of $U^2$ are also elements of $U$ and it has rank $n$.
I think what confused you is thinking that the things in the proof were intended to prove $text{rank}(U^2)=text{rank}(U)$, but in reality that's known from this argument via basic algebra. The real reason it's in there is because you want to trap the group $RV$ which is between them to compute its rank, which is part of what was being shown.
answered Jul 17 '14 at 0:04
Adam HughesAdam Hughes
32.1k83670
$endgroup$
$begingroup$
"But then if the subgroup is of finite index, they have the same rank." - is this a general fact about free abelian groups?
$endgroup$
– user164793
Jul 17 '14 at 0:20
$begingroup$
@user164793 Finitely generated ones? Yes. Easier way than my previous comment: since you're dealing with a PID, you can form the vector space spanned by the basis elements for $Bbb Z^n$, and then notice that modding out by a subspace of the same rank gives the trivial vector space by the definition of dimension.
$endgroup$
– Adam Hughes
Jul 17 '14 at 0:33
$begingroup$
I'm sorry, I still don't follow why the finite index of $U^2$ in $U$ allows us to conclude they are of equal rank. I'm fairly weak in algebra, if you haven't picked up on that :( I understand the rest.
$endgroup$
– user164793
Jul 17 '14 at 0:41
$begingroup$
@user164793 give me a moment to edit my post, I'll just do the proof, it's not hard.
$endgroup$
– Adam Hughes
Jul 17 '14 at 0:43
$begingroup$
Thank you very much, I understand now. I really appreciate you taking the time.
$endgroup$
– user164793
Jul 17 '14 at 0:49
|
show 2 more comments
$begingroup$
All right, well first of all Dirichlet's unit theorem tells you that the unit group (modulo roots of unity, which are irrelevant for computing rank) is a finitely generated free group on $r+s-1$ generators with $r$ the number of real embeddings and $s$ the number of complex ones.
If we use additive notation
$$U/text{Tor}(U)cong bigoplus_{i=1}^{r+s-1}Bbb Z$$
it is easier to see that the squares here are a subgroup of index $2^{r+s-1}$ since $U^2/text{Tor}(U^2)$ is generated by squares of generators of the torsion-free part of $U$. But then if the subgroup is of finite index, they have the same rank.
Edit: In case you've not seen the proof of finite index explicitly, it's quite direct:
If $xin U^2$ and $x_1,ldots, x_{r+s-1}$ are generators for $U/text{Tor}(U)$ then $x=(x_1^{e_1}ldots x_{r+s-1}^{e_{r+s-1}})^2$ so that the powers are all even, hence $x_1^2,ldots, x_{r+s-1}^2$ generate $U^2/text{Tor}(U^2)$. Consider the projection map
$$U/text{Tor}(U)to U^2/text{Tor}(U^2)$$
which is given on the additive structure by the multiplication by $2$ map, you get an isomorphism
$$U^2/text{Tor}(U^2)cong (2Bbb Z)^{r+s-1}$$
for the image.
Edit 2: Let's just do this explicitly. In this particular case it's not even hard. You can see that $x_1^{2e_1}cdotldotscdot x_{r+s-1}^{2e_{r+s-1}}=1$ (i.e. the identity element of the group) implies that all the $e_i$ are zero because we know that this holds in $U$ which is a bigger group. Hence $x_1,ldots ,x_{r+s-1}$ are linearly independent elements of $U^2$ hence it has rank $ge r+s-1$, but it also has rank $le r+s-1$ because all elements of $U^2$ are also elements of $U$ and it has rank $n$.
I think what confused you is thinking that the things in the proof were intended to prove $text{rank}(U^2)=text{rank}(U)$, but in reality that's known from this argument via basic algebra. The real reason it's in there is because you want to trap the group $RV$ which is between them to compute its rank, which is part of what was being shown.
answered Jul 17 '14 at 0:04
Adam HughesAdam Hughes
32.1k83670
$endgroup$
$begingroup$
"But then if the subgroup is of finite index, they have the same rank." - is this a general fact about free abelian groups?
$endgroup$
– user164793
Jul 17 '14 at 0:20
$begingroup$
@user164793 Finitely generated ones? Yes. Easier way than my previous comment: since you're dealing with a PID, you can form the vector space spanned by the basis elements for $Bbb Z^n$, and then notice that modding out by a subspace of the same rank gives the trivial vector space by the definition of dimension.
$endgroup$
– Adam Hughes
Jul 17 '14 at 0:33
$begingroup$
I'm sorry, I still don't follow why the finite index of $U^2$ in $U$ allows us to conclude they are of equal rank. I'm fairly weak in algebra, if you haven't picked up on that :( I understand the rest.
$endgroup$
– user164793
Jul 17 '14 at 0:41
$begingroup$
@user164793 give me a moment to edit my post, I'll just do the proof, it's not hard.
$endgroup$
– Adam Hughes
Jul 17 '14 at 0:43
$begingroup$
Thank you very much, I understand now. I really appreciate you taking the time.
$endgroup$
– user164793
Jul 17 '14 at 0:49
|
show 2 more comments
$begingroup$
All right, well first of all Dirichlet's unit theorem tells you that the unit group (modulo roots of unity, which are irrelevant for computing rank) is a finitely generated free group on $r+s-1$ generators with $r$ the number of real embeddings and $s$ the number of complex ones.
If we use additive notation
$$U/text{Tor}(U)cong bigoplus_{i=1}^{r+s-1}Bbb Z$$
it is easier to see that the squares here are a subgroup of index $2^{r+s-1}$ since $U^2/text{Tor}(U^2)$ is generated by squares of generators of the torsion-free part of $U$. But then if the subgroup is of finite index, they have the same rank.
Edit: In case you've not seen the proof of finite index explicitly, it's quite direct:
If $xin U^2$ and $x_1,ldots, x_{r+s-1}$ are generators for $U/text{Tor}(U)$ then $x=(x_1^{e_1}ldots x_{r+s-1}^{e_{r+s-1}})^2$ so that the powers are all even, hence $x_1^2,ldots, x_{r+s-1}^2$ generate $U^2/text{Tor}(U^2)$. Consider the projection map
$$U/text{Tor}(U)to U^2/text{Tor}(U^2)$$
which is given on the additive structure by the multiplication by $2$ map, you get an isomorphism
$$U^2/text{Tor}(U^2)cong (2Bbb Z)^{r+s-1}$$
for the image.
Edit 2: Let's just do this explicitly. In this particular case it's not even hard. You can see that $x_1^{2e_1}cdotldotscdot x_{r+s-1}^{2e_{r+s-1}}=1$ (i.e. the identity element of the group) implies that all the $e_i$ are zero because we know that this holds in $U$ which is a bigger group. Hence $x_1,ldots ,x_{r+s-1}$ are linearly independent elements of $U^2$ hence it has rank $ge r+s-1$, but it also has rank $le r+s-1$ because all elements of $U^2$ are also elements of $U$ and it has rank $n$.
I think what confused you is thinking that the things in the proof were intended to prove $text{rank}(U^2)=text{rank}(U)$, but in reality that's known from this argument via basic algebra. The real reason it's in there is because you want to trap the group $RV$ which is between them to compute its rank, which is part of what was being shown.
answered Jul 17 '14 at 0:04
Adam HughesAdam Hughes
32.1k83670
$endgroup$
All right, well first of all Dirichlet's unit theorem tells you that the unit group (modulo roots of unity, which are irrelevant for computing rank) is a finitely generated free group on $r+s-1$ generators with $r$ the number of real embeddings and $s$ the number of complex ones.
If we use additive notation
$$U/text{Tor}(U)cong bigoplus_{i=1}^{r+s-1}Bbb Z$$
it is easier to see that the squares here are a subgroup of index $2^{r+s-1}$ since $U^2/text{Tor}(U^2)$ is generated by squares of generators of the torsion-free part of $U$. But then if the subgroup is of finite index, they have the same rank.
Edit: In case you've not seen the proof of finite index explicitly, it's quite direct:
If $xin U^2$ and $x_1,ldots, x_{r+s-1}$ are generators for $U/text{Tor}(U)$ then $x=(x_1^{e_1}ldots x_{r+s-1}^{e_{r+s-1}})^2$ so that the powers are all even, hence $x_1^2,ldots, x_{r+s-1}^2$ generate $U^2/text{Tor}(U^2)$. Consider the projection map
$$U/text{Tor}(U)to U^2/text{Tor}(U^2)$$
which is given on the additive structure by the multiplication by $2$ map, you get an isomorphism
$$U^2/text{Tor}(U^2)cong (2Bbb Z)^{r+s-1}$$
for the image.
Edit 2: Let's just do this explicitly. In this particular case it's not even hard. You can see that $x_1^{2e_1}cdotldotscdot x_{r+s-1}^{2e_{r+s-1}}=1$ (i.e. the identity element of the group) implies that all the $e_i$ are zero because we know that this holds in $U$ which is a bigger group. Hence $x_1,ldots ,x_{r+s-1}$ are linearly independent elements of $U^2$ hence it has rank $ge r+s-1$, but it also has rank $le r+s-1$ because all elements of $U^2$ are also elements of $U$ and it has rank $n$.
I think what confused you is thinking that the things in the proof were intended to prove $text{rank}(U^2)=text{rank}(U)$, but in reality that's known from this argument via basic algebra. The real reason it's in there is because you want to trap the group $RV$ which is between them to compute its rank, which is part of what was being shown.
answered Jul 17 '14 at 0:04
Adam HughesAdam Hughes
32.1k83670
edited Jan 2 at 16:49
answered Jul 17 '14 at 0:04
Adam HughesAdam Hughes
32.1k83670
answered Jul 17 '14 at 0:04
Adam HughesAdam Hughes
32.1k83670
answered Jul 17 '14 at 0:04
Adam HughesAdam Hughes
32.1k83670
32.1k83670
$begingroup$
"But then if the subgroup is of finite index, they have the same rank." - is this a general fact about free abelian groups?
$endgroup$
– user164793
Jul 17 '14 at 0:20
$begingroup$
@user164793 Finitely generated ones? Yes. Easier way than my previous comment: since you're dealing with a PID, you can form the vector space spanned by the basis elements for $Bbb Z^n$, and then notice that modding out by a subspace of the same rank gives the trivial vector space by the definition of dimension.
$endgroup$
– Adam Hughes
Jul 17 '14 at 0:33
$begingroup$
I'm sorry, I still don't follow why the finite index of $U^2$ in $U$ allows us to conclude they are of equal rank. I'm fairly weak in algebra, if you haven't picked up on that :( I understand the rest.
$endgroup$
– user164793
Jul 17 '14 at 0:41
$begingroup$
@user164793 give me a moment to edit my post, I'll just do the proof, it's not hard.
$endgroup$
– Adam Hughes
Jul 17 '14 at 0:43
$begingroup$
Thank you very much, I understand now. I really appreciate you taking the time.
$endgroup$
– user164793
Jul 17 '14 at 0:49
|
show 2 more comments
$begingroup$
"But then if the subgroup is of finite index, they have the same rank." - is this a general fact about free abelian groups?
$endgroup$
– user164793
Jul 17 '14 at 0:20
$begingroup$
@user164793 Finitely generated ones? Yes. Easier way than my previous comment: since you're dealing with a PID, you can form the vector space spanned by the basis elements for $Bbb Z^n$, and then notice that modding out by a subspace of the same rank gives the trivial vector space by the definition of dimension.
$endgroup$
– Adam Hughes
Jul 17 '14 at 0:33
$begingroup$
I'm sorry, I still don't follow why the finite index of $U^2$ in $U$ allows us to conclude they are of equal rank. I'm fairly weak in algebra, if you haven't picked up on that :( I understand the rest.
$endgroup$
– user164793
Jul 17 '14 at 0:41
$begingroup$
@user164793 give me a moment to edit my post, I'll just do the proof, it's not hard.
$endgroup$
– Adam Hughes
Jul 17 '14 at 0:43
$begingroup$
Thank you very much, I understand now. I really appreciate you taking the time.
$endgroup$
– user164793
Jul 17 '14 at 0:49
$begingroup$
"But then if the subgroup is of finite index, they have the same rank." - is this a general fact about free abelian groups?
$endgroup$
– user164793
Jul 17 '14 at 0:20
$begingroup$
"But then if the subgroup is of finite index, they have the same rank." - is this a general fact about free abelian groups?
$endgroup$
– user164793
Jul 17 '14 at 0:20
$begingroup$
@user164793 Finitely generated ones? Yes. Easier way than my previous comment: since you're dealing with a PID, you can form the vector space spanned by the basis elements for $Bbb Z^n$, and then notice that modding out by a subspace of the same rank gives the trivial vector space by the definition of dimension.
$endgroup$
– Adam Hughes
Jul 17 '14 at 0:33
$begingroup$
@user164793 Finitely generated ones? Yes. Easier way than my previous comment: since you're dealing with a PID, you can form the vector space spanned by the basis elements for $Bbb Z^n$, and then notice that modding out by a subspace of the same rank gives the trivial vector space by the definition of dimension.
$endgroup$
– Adam Hughes
Jul 17 '14 at 0:33
$begingroup$
I'm sorry, I still don't follow why the finite index of $U^2$ in $U$ allows us to conclude they are of equal rank. I'm fairly weak in algebra, if you haven't picked up on that :( I understand the rest.
$endgroup$
– user164793
Jul 17 '14 at 0:41
$begingroup$
I'm sorry, I still don't follow why the finite index of $U^2$ in $U$ allows us to conclude they are of equal rank. I'm fairly weak in algebra, if you haven't picked up on that :( I understand the rest.
$endgroup$
– user164793
Jul 17 '14 at 0:41
$begingroup$
@user164793 give me a moment to edit my post, I'll just do the proof, it's not hard.
$endgroup$
– Adam Hughes
Jul 17 '14 at 0:43
$begingroup$
@user164793 give me a moment to edit my post, I'll just do the proof, it's not hard.
$endgroup$
– Adam Hughes
Jul 17 '14 at 0:43
$begingroup$
Thank you very much, I understand now. I really appreciate you taking the time.
$endgroup$
– user164793
Jul 17 '14 at 0:49
$begingroup$
Thank you very much, I understand now. I really appreciate you taking the time.
$endgroup$
– user164793
Jul 17 '14 at 0:49
|
show 2 more comments
$begingroup$
$U/U^2$ is a finitely generated (multiplicative) abelian group every element of which has order $2$. Thus it is finite.
answered Jul 17 '14 at 0:04
Rene SchipperusRene Schipperus
32.2k11960
$endgroup$
add a comment |
$begingroup$
$U/U^2$ is a finitely generated (multiplicative) abelian group every element of which has order $2$. Thus it is finite.
answered Jul 17 '14 at 0:04
Rene SchipperusRene Schipperus
32.2k11960
$endgroup$
add a comment |
$begingroup$
$U/U^2$ is a finitely generated (multiplicative) abelian group every element of which has order $2$. Thus it is finite.
answered Jul 17 '14 at 0:04
Rene SchipperusRene Schipperus
32.2k11960
$endgroup$
$U/U^2$ is a finitely generated (multiplicative) abelian group every element of which has order $2$. Thus it is finite.
answered Jul 17 '14 at 0:04
Rene SchipperusRene Schipperus
32.2k11960
answered Jul 17 '14 at 0:04
Rene SchipperusRene Schipperus
32.2k11960
answered Jul 17 '14 at 0:04
Rene SchipperusRene Schipperus
32.2k11960
answered Jul 17 '14 at 0:04
Rene SchipperusRene Schipperus
32.2k11960
32.2k11960
add a comment |
add a comment |
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$begingroup$
I cannot access the paper from home: can you type the theorem? I suspect it is a finite index thing, depending on how you're defining $U^2$ (some do products, others do squares)
$endgroup$
– Adam Hughes
Jul 16 '14 at 23:52
$begingroup$
I believe $U^2$ means squares but it is not actually stated. It just seems that way from the proof.
$endgroup$
– user164793
Jul 17 '14 at 0:03
$begingroup$
It does indeed, I can tell from the proof. I've posted an answer.
$endgroup$
– Adam Hughes
Jul 17 '14 at 0:05